Proof by strong induction combinatorics problem: $1(1!) + 2(2!) + 3(3!) + \dots + n(n!) = (n+1)! - 1$

Question

$1(1!) + 2(2!) + 3(3!) + \dots + n(n!) = (n+1)! - 1$

How do we prove this by strong induction?

I know how to do it with weak induction, but how would strong induction work with this problem?

Answer 1

The induction step $$(n+1)(n+1)!+(n+1)!=(n+2)!$$